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In this weeks discussion, which is our final week, we are learning about factoring. When we

factor an expression, we write the expression as a product. To describe it another way we are

splitting an expression into a multiplication of simpler expressions. We have been asked to

complete the exercises on pages 345-346, and 353. Using the third letter in my last name, which

is O I am assigned to numbers 72 and 104 on page 346 and number 38 on page 353.

For number 72 I am to factor each polynomial completely using the methods discussed in

Sections 5.1 through 5.3 of our textbook.

-a3-100a I will factorout the GCFof -a from the expression -a

3

-a(a2)-a(100) Next I factor out the GCF of -a from the expression -100a

-a(a2+ 100) I then factorout the GCFof -a from -a

3100a and arrive at my final

answer

In my next exercise, which is number, 104 I am factoring completely where the GCF has avariable and further factors to perfect square trinomial.

-4w3-16w

2+20w I factorout the GCFof 4 from the expression -64

-4(-16)-16w2+20w Next I factorout the GCFof 4 from the expression 16w

2

4(-16)+4(-4w2)+20w I then factorout the GCFof 4 from the expression 20w

4(-16)+4(-4w2)+4(5w)Finally I factorout the GCFof 4 from -64,16w

2, and 20w

where I arrive at my final answer.

My final exercise uses the ac Method where the leading coefficient only has multiple factors.

On page 353 of our textbook I am assigned number 38.

15x2-7x-2 In order to solve by groupingusing ac which is (-30) and the

sum is b(-7)

a = 15, b = -7, c =-2 I then rewrite the middle term as a sum of two terms

15(-2)=-30 The sum of this is -7

15x2-10x+3x-2 This is my final answer to the ac grouping

((15x2-10x)+(3x-2)) The two terms are now groupedtogether with the second two

terms

(5x(3x-2)+3x-2) The GCFis factoredfrom each group

(5x(3x-2)+3x-2) Both groups are grouped together and both contain the factor(3x-

2)

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(5x+1) (3x-2) I arrive at my final answer